The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M . Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its center. The magnitudes of the acceleration experienced by the blocks is _ _ _ _ (assume no slipping of string on pul

Step-by-step solution

The pulley consists of a rim (mass M ) and two rods (each mass M ) forming a diameter structure. The moment of inertia is: I = I_ rim + I_ rods = MR^2 + 2 MR^2 3 = 5MR^2 3 . For the Atwood pulley system, applying Newton's second law: Mass m (upward): T₁ = m(g-a) Mass M (downward): T₂ = M(g+a) Pulley rotation: T₂ R - T₁ R = I = 5MR^2 3 a R Thus: T₂ - T₁ = 5Ma 3 Substituting: M(g+a) - m(g-a) = 5Ma 3 (M-m)g + (M+m)a = 5Ma 3 (M-m)g = a ( 5M 3 - M - m ) = a ( 2M 3 - m ) a = (M-m)g 2M-3m 3 = 3(M-m)g 2M+3m , which is opti